Physics - Center of Mass Question with Solution | TestHub

From conservation of linear momentum,
Along $\mathrm{x}$-axis:

${\text{10Mcos30}}^0+5\left(\text{2M}cos{45}^0\right)=\text{M}{v}_{1}cos{30}^0+2\text{M}{v}_{2}cos{45}^0$
$10\mathrm{M}\frac{\sqrt{3}}{2}+10\mathrm{M}\frac{1}{\sqrt{2}}=\mathrm{M}\frac{\sqrt{3}}{2}{\mathrm{v}}_1+2\mathrm{M}{\mathrm{v}}_2\frac{1}{\sqrt{2}}$
$\frac{\sqrt{3}}{2}{\mathrm{v}}_1+\sqrt{2}{\mathrm{v}}_2=5(\sqrt{3}+\sqrt{2})$ … (1)
Along $\mathrm{y}$-axis:
$-10\mathrm{M}\frac{1}{2}+10\mathrm{M}\frac{1}{\sqrt{2}}=\mathrm{M}{\mathrm{v}}_1\frac{1}{2}-2\mathrm{M}{\mathrm{v}}_2\frac{1}{\sqrt{2}}$
$\frac{{\mathrm{v}}_1}{2}-\sqrt{2}{\mathrm{v}}_2=5(\sqrt{2}-1)$ … (2)
From equations (1) and (2)
${\mathrm{v}}_1\left(\frac{\sqrt{3}}{2}+\frac{1}{2}\right)=5\left(\sqrt{3}+\sqrt{2}\right)+5(\sqrt{2}-1)$
${\mathrm{v}}_1=13.02\mathrm{m}/\mathrm{s}$
${\mathrm{v}}_2=\frac{5\sqrt{3}+5\sqrt{2}-13.02\times \frac{\sqrt{3}}{2}}{\sqrt{2}}$
$=19.72\mathrm{m} {\mathrm{s}}^{-1}$